Differential cross-sectional area - Rutherford's experiment

[mw98]

I'm struggling to understand the importance of the differential cross-sectional area in Rutherford's scattering experiment, dσ/dθ. In one part of my course notes it is explained as 'the number of scatterings between θ and θ + dθ per unit flux, per unit range of angle'. However, dσ itself is described as the 'infinitesimal effective area for collision'. The graph of dσ/dθ shows a large reduction in the number of scatterings as the scattering angle is increased, but only for small angles. At larger θ, increasing θ has very little effect on the number of scatterings.

Am I understanding the results correctly? Is the 'effective area for collision' unaffected by change of scattering angle, only the number of scatterings is affected? If so, why is it called the 'differential cross-sectional area' when the number of observed scatterings is the only thing that changes, not the actual effective area for collision?

In my notes it is stated: 'Although the differential cross-section falls rapidly with the scattering angle, the cross-section at large angles is still much larger than would have been obtained from Thomson’s ‘current cake’ model of the atom in which electrons are embedded in a ‘dough’ of positive charge.'

Why is this?

 

[Charles Link]

Differential cross section can be understood as a mapping of area $\sigma$ in the plane of the scatterer to a region of the solid angle $\Omega$ to which it gets scattered. (This assumes a one-to-one mapping, although some scattering could be a random and average result).$\\$ If $P$ is the total incident power contained in an area $\sigma$ of the scattering plane, that same amount of power $P$ can also be a function of the corresponding sold angle $\Omega$. $\\$ If we let $P(\Omega)=$ total integrated scattered power for solid angle less than or equal to $\Omega$, $\\$ and $P(\sigma)=$ total integrated power for area less than $\sigma$, $\\$ then the differential scattering cross section is $\frac{d \sigma}{d \Omega}=\frac{(\frac{dP(\Omega)}{d \Omega})}{(\frac{d P(\sigma)}{d \sigma})} =\frac{I}{E}$, $\\$ where the $P(\Omega)=P(\sigma)$ and cancels in numerator and denominator. (And notice $d \sigma$ winds up in the numerator). $\\$ The intensity $I$ is the scattered intensity (watts/steradian) at some angle $(\theta, \phi)$, and the irradiance $E$ is the power per unit area incident (watts/m^2) on the target. $\\$ The expressions on the right side of the above equation(s) are much more readily understood, than just trying to make sense of looking at the term $\frac{d \sigma}{d \Omega}$. $\\$ The differential scattering cross section can be integrated over all scattering angles to get the total size of the scatterer: $\sigma_{total}=\int (\frac{d \sigma}{d \Omega} ) \, d \Omega$, where the unscattered power is not considered in this calculation.$\\$ [If the incident power is far from the target(s),(e.g. if it goes around the foil, and also if it goes into the foil but travels in the open spaces between the gold atoms), it travels in a straight line to the receiver, and is unscattered]. $\\$ Experimentally, it is usually fairly easy to determine what belongs in the category of unscattered power. $\\$ In the case of Rutherford's experiment, the thin gold foil contains many, many scatterers $N$. From an analysis of the incident irradiance $E$and scattered intensity $I(\theta, \phi)$, the total scattering area $\sigma_{total}$ can be computed. $\\$ (This area $\sigma_{total}$ is many,many times smaller than the area of the foil. Most of the foil is empty space as far as scattering is concerned. The electrons are not massive enough to do any scattering. If the positive nuclei had been spead out like the electrons are, they also would be incapable of doing appreciable scattering, and all of the incident power would wind up in the unscattered category, and would be ignored in performing the calculation of $\sigma_{total}$). $\\$ From that, and knowing how many atoms $N$ are in the foil, the scattering cross section $\sigma_{single}=\frac{\sigma_{total}}{N}$ can be computed/estimated. $\\$ Additional item: For some scatterers, e.g. small "bee-bees" off of a large smooth spherical target, by using angle of incidence=angle of reflection, the mathematical relationship between $\sigma$ and $\Omega$, and thereby $\frac{d \sigma}{d \Omega}$ can be computed. For other scatterers, such as Rutherford's gold foil of atomic scatterers, the function $\frac{d \sigma}{d \Omega }$ is determined experimentally. Note that $\frac{d \sigma}{d \Omega}$ is basically a function of $(\theta, \phi)$ in a spherical coordinate system, and it is integrated over $d \Omega=\sin{\theta} \, d \theta \, d \phi$, i.e. the experimentally determined $\frac{d \sigma}{d \Omega}$ is numerically integrated, [using the sampling of the data points of the scattered intensity $I=I(\theta, \phi)$ over the entire $4 \pi$ steradians], over $d \Omega$ to experimentally determine $\sigma_{total}$. $\\$ Additional item: If the experimental scenario was to scatter bee-bees off of two cone-shaped targets that had the same projected areas (i.e. area of the circular base is same for both) but had different angles forming the cone, the $\sigma_{total}$(= area of base) that is computed experimentally would be the same for both even though the scattering would be different. The cones would scatter the bee-bees into a ring in both cases, with the $\theta$ (location of the ring) being different in the two cases of where the peak intensity occurrs. Thereby knowing $\sigma_{total}$ does not tell us everything about the scatterer. $\\$ For both cones the total scattered power $P=\int I \, d \Omega =E \, A_{base}$. Thereby $\sigma_{total}=\frac{\int I \, d \Omega}{E}=A_{base}$. $\\$ (In the case of bee-bees, the power $P$ would be replaced by the number of particles(bee-bees) per second. And similarly for alpha particles incident on a gold foil. ). $\\$ Notice for each of these cones that we must have $\frac{d \sigma}{d \Omega}=\frac{A_{base}}{2 \pi \sin{\theta}} \delta(\theta-\theta_o)$, where $\theta_o$ will be different for each cone. (In both cases, $\theta_o$ is equal to the full cone angle for the given cone, where we have assumed angle of incidence=angle of reflection for the scattering.). $\\$ Edit: Additional item: The differential scattering cross section $\frac{d \sigma}{d \Omega}$ can be somewhat confusing until it is shown for what it is (as seen above) as $\frac{d \sigma}{d \Omega}= \frac{I}{E}$. What it basically says, (and this should be quite logical), is that the total scattering cross section depends simply on the total scattered power: $\sigma_{total} =\frac{P_{scattered}}{E}$, where the total scattered power is simply $P_{scattered}=\int I \, d \Omega$. $\\$ For the case of the Rayleigh scattering, if the target was a positive charge that was uniformly distributed, the power would all wind up in the unscattered category. The uniform density of positive charge would not scatter any of the incident beam appreciably, so that even though the same amount of matter is present in both cases, it is only when the positive charge is grouped as a bunch of small nuclei that there is any (measurable) scattered power. The total scattering cross section $\sigma_{total}$ is still a very small area, (even in the case of the many millions and millions of nuclei doing the scattering=(most of the gold foil consists of empty space for which the alpha particles are basically unscattered)), but because there is a finite $P_{scattered} =\int I \, d \Omega$, the scattering cross section, $\sigma_{total}$, (and thereby even the cross section for a single atom), can be computed from the measurements of $I=I(\theta, \phi)$.

 

[mw98]

Differential cross section can be understood as a mapping of area $\sigma$ in the plane of the scatterer to a region of the solid angle $\Omega$ to which it gets scattered. (This assumes a one-to-one mapping, although some scattering could be a random and average result).$\\$ If $P$ is the total incident power contained in an area $\sigma$ of the scattering plane, that same amount of power $P$ can also be a function of the corresponding sold angle $\Omega$. $\\$ If we let $P(\Omega)=$ total integrated scattered power for solid angle less than or equal to $\Omega$, $\\$ and $P(\sigma)=$ total integrated power for area less than $\sigma$, $\\$ then the differential scattering cross section is $\frac{d \sigma}{d \Omega}=\frac{(\frac{dP(\Omega)}{d \Omega})}{(\frac{d P(\sigma)}{d \sigma})} =\frac{I}{E}$, $\\$ where the $P(\Omega)=P(\sigma)$ and cancels in numerator and denominator. (And notice $d \sigma$ winds up in the numerator). $\\$ The intensity $I$ is the scattered intensity (watts/steradian) at some angle $(\theta, \phi)$, and the irradiance $E$ is the power per unit area incident (watts/m^2) on the target. $\\$ The expressions on the right side of the above equation(s) are much more readily understood, than just trying to make sense of looking at the term $\frac{d \sigma}{d \Omega}$. $\\$ The differential scattering cross section can be integrated over all scattering angles to get the total size of the scatterer: $\sigma_{total}=\int (\frac{d \sigma}{d \Omega} ) \, d \Omega$, where the unscattered power is not considered in this calculation.$\\$ [If the incident power is far from the target(s),(e.g. if it goes around the foil, and also if it goes into the foil but travels in the open spaces between the gold atoms), it travels in a straight line to the receiver, and is unscattered]. $\\$ Experimentally, it is usually fairly easy to determine what belongs in the category of unscattered power. $\\$ In the case of Rutherford's experiment, the thin gold foil contains many, many scatterers $N$. From an analysis of the incident irradiance $E$and scattered intensity $I(\theta, \phi)$, the total scattering area $\sigma_{total}$ can be computed. $\\$ (This area $\sigma_{total}$ is many,many times smaller than the area of the foil. Most of the foil is empty space as far as scattering is concerned. The electrons are not massive enough to do any scattering. If the positive nuclei had been spead out like the electrons are, they also would be incapable of doing appreciable scattering, and all of the incident power would wind up in the unscattered category, and would be ignored in performing the calculation of $\sigma_{total}$). $\\$ From that, and knowing how many atoms $N$ are in the foil, the scattering cross section $\sigma_{single}=\frac{\sigma_{total}}{N}$ can be computed/estimated. $\\$ Additional item: For some scatterers, e.g. small "bee-bees" off of a large smooth spherical target, by using angle of incidence=angle of reflection, the mathematical relationship between $\sigma$ and $\Omega$, and thereby $\frac{d \sigma}{d \Omega}$ can be computed. For other scatterers, such as Rutherford's gold foil of atomic scatterers, the function $\frac{d \sigma}{d \Omega }$ is determined experimentally. Note that $\frac{d \sigma}{d \Omega}$ is basically a function of $(\theta, \phi)$ in a spherical coordinate system, and it is integrated over $d \Omega=\sin{\theta} \, d \theta \, d \phi$, i.e. the experimentally determined $\frac{d \sigma}{d \Omega}$ is numerically integrated, [using the sampling of the data points of the scattered intensity $I=I(\theta, \phi)$ over the entire $4 \pi$ steradians], over $d \Omega$ to experimentally determine $\sigma_{total}$. $\\$ Additional item: If the experimental scenario was to scatter bee-bees off of two cone-shaped targets that had the same projected areas (i.e. area of the circular base is same for both) but had different angles forming the cone, the $\sigma_{total}$(= area of base) that is computed experimentally would be the same for both even though the scattering would be different. The cones would scatter the bee-bees into a ring in both cases, with the $\theta$ (location of the ring) being different in the two cases of where the peak intensity occurrs. Thereby knowing $\sigma_{total}$ does not tell us everything about the scatterer. $\\$ For both cones the total scattered power $P=\int I \, d \Omega =E \, A_{base}$. Thereby $\sigma_{total}=\frac{\int I \, d \Omega}{E}=A_{base}$. $\\$ (In the case of bee-bees, the power $P$ would be replaced by the number of particles(bee-bees) per second. And similarly for alpha particles incident on a gold foil. ). $\\$ Notice for each of these cones that we must have $\frac{d \sigma}{d \Omega}=\frac{A_{base}}{2 \pi \sin{\theta}} \delta(\theta-\theta_o)$, where $\theta_o$ will be different for each cone. (In both cases, $\theta_o$ is equal to the full cone angle for the given cone, where we have assumed angle of incidence=angle of reflection for the scattering.). $\\$ Edit: Additional item: The differential scattering cross section $\frac{d \sigma}{d \Omega}$ can be somewhat confusing until it is shown for what it is (as seen above) as $\frac{d \sigma}{d \Omega}= \frac{I}{E}$. What it basically says, (and this should be quite logical), is that the total scattering cross section depends simply on the total scattered power: $\sigma_{total} =\frac{P_{scattered}}{E}$, where the total scattered power is simply $P_{scattered}=\int I \, d \Omega$. $\\$ For the case of the Rayleigh scattering, if the target was a positive charge that was uniformly distributed, the power would all wind up in the unscattered category. The uniform density of positive charge would not scatter any of the incident beam appreciably, so that even though the same amount of matter is present in both cases, it is only when the positive charge is grouped as a bunch of small nuclei that there is any (measurable) scattered power. The total scattering cross section $\sigma_{total}$ is still a very small area, (even in the case of the many millions and millions of nuclei doing the scattering=(most of the gold foil consists of empty space for which the alpha particles are basically unscattered)), but because there is a finite $P_{scattered} =\int I \, d \Omega$, the scattering cross section, $\sigma_{total}$, (and thereby even the cross section for a single atom), can be computed from the measurements of $I=I(\theta, \phi)$.

Thank you so much. Sorry for the late reply, this has cleared up so much for me.

 

[Charles Link]

Thank you so much. Sorry for the late reply, this has cleared up so much for me.

I think many, if not most textbook treatments on this subject are rather incomplete. The explanation I presented above is what I figured out from the bits and pieces that my classmates and I were given. It was only many years later, with a couple of concepts that I figured out over time, that I was able to put together a thorough treatment of it. I have yet to see a clear treatment of this topic in a textbook.